Differential Equations 1 Question 15
15. Let $f: R \rightarrow R$ be a differentiable function with $f(0)=0$. If $y=f(x)$ satisfies the differential equation $\frac{d y}{d x}=(2+5 y)(5 y-2)$, then the value of $\lim _{x \rightarrow-\infty} f(x)$ is $\left.\ldots \ldots . . ..\right)$
Assertion and Reason
For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I.
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
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Answer:
Correct Answer: 15. (0)
Solution:
- We have,
$$ \begin{aligned} & \quad \frac{d y}{d x}=(2+5 y)(5 y-2) \\ & \Rightarrow \quad \frac{d y}{25 y^{2}-4}=d x \Rightarrow \frac{1}{25} \frac{d y}{y^{2}-\frac{4}{25}}=d x \end{aligned} $$
On integrating both sides, we get
$$ \begin{aligned} & \frac{1}{25} \int \frac{d y}{y^{2}-\frac{2}{5}}=\int d x \\ \Rightarrow & \frac{1}{25} \times \frac{1}{2 \times \frac{2}{5}} \log \left|\frac{y-2 / 5}{y+2 / 5}\right|=x+C \\ \Rightarrow & \log \left|\frac{5 y-2}{5 y+2}\right|=20(x+C) \\ \Rightarrow & \left|\frac{5 y-2}{5 y+2}\right|=A e^{20 x}\left[\because e^{20 C}=A\right] \end{aligned} $$
when $x=0 \Rightarrow y=0$, then $A=1$
$\therefore\left|\frac{5 y-2}{5 y+2}\right|=e^{20 x}$
$$ \lim _{x \rightarrow-\infty}\left|\frac{5 f(x)-2}{5 f(x)+2}\right|=\lim _{x \rightarrow-\infty} e^{20 x} $$
$$ \begin{aligned} & \Rightarrow \quad \lim _{n \rightarrow-\infty}\left|\frac{5 f(x)-2}{5 f(x)+2}\right|=0 \\ & \Rightarrow \quad \lim _{n \rightarrow-\infty} 5 f(x)-2=0 \\ & \Rightarrow \quad \lim _{n \rightarrow-\infty} f(x)=\frac{2}{5}=0.4 \end{aligned} $$
